Pick Game

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Problem Description

　　One day, hh plays a game with GL. On a n*m matrix, each gird has a value. The player could only choose the gird that is adjacent to at least two empty girds (A grid outside the matrix also regard as empty). Adjacent means two girds share a common edge. If one play chooses one gird, he will get the value and the gird will be empty. They play in turn.
　　GL and hh are clever boys. They will choose the best strategy, hh plays first, and he wants to know the maximal value he could get.

Input

On the first line of input is a single positive integer, 1<=T<=50, specifying the number of test cases to follow.
Each test case begins with two numbers n and m. (2<=n, m<=5)
Then n lines follow and each lines with m numbers Vij. (0< Vij <=1000)

Output

Output the maximal value hh could get.

Sample Input

12 29 87 6

Sample Output

Source

2014-03-PK2

题意：n*m的矩阵，2个人轮流取一个格子的值，要求取的格子四周至少有2个空的格子，hh先取，求他能取得的最大值

分析：记忆化搜索，dp表示当前状态hh比GL多的值，0表示这一个未取，1表示已取，一共最多有2^25种状态，用hash散列来存

#include <iostream>#include<cstdio>#include<cstring>#include<string>#include<algorithm>#define mod 10007#define N 10010#include<vector>using namespace std;vector<int> v[N],dp[N];int n,m,cnt,Map[6][6];int dir[4][2]={{0,1},{0,-1},{1,0},{-1,0}};bool ok(int x,int y){    if(x<n&&x>=0&&y<m&&y>=0)    return true;    return false;}int Gets(int st){    int ret=0;    for(int i=0;i<cnt;i++){        if(((1<<i)&st)==0){            ret+=Map[i/m][i%m];        }    }    return ret;}int cal(int st){    int ret=0,i;    for(i=0;i<cnt;i++)    if((1<<i)&st)    ret++;    return ret;}int dfs(int st){    int next,ans,i,j,tmp,x,y,xx,yy;    x=st%mod;    for(y=v[x].size()-1;y>=0;y--){        if(v[x][y]==st)        return dp[x][y];    }    if(cal(st)==cnt-1)    return Gets(st);    ans=-20006;    for(i=0;i<cnt;i++){        if(((1<<i)&st)==0){            x=i/m;            y=i%m;            tmp=0;            for(j=0;j<4;j++){                xx=x+dir[j][0];                yy=y+dir[j][1];                if(!ok(xx,yy)||(ok(xx,yy)&&(((1<<(xx*m+yy))&st)))){                    tmp++;                }            }            if(tmp>=2){                next=(1<<i) | st;                ans=max(ans,Map[x][y]-dfs(next));            }        }    }    v[st%mod].push_back(st);    dp[st%mod].push_back(ans);    return ans;}int main(){    int t,i,j;    scanf("%d",&t);    while(t--){        for(i=0;i<mod;i++){            dp[i].clear();            v[i].clear();        }        scanf("%d%d",&n,&m);        for(i=0;i<n;i++){            for(j=0;j<m;j++){                scanf("%d",&Map[i][j]);            }        }        cnt=n*m;        printf("%d\n",(dfs(0)+Gets(0))/2);    }    return 0;}

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